getchar() just doesn't get it.

What is wrong with the following code, that getchar() won't wait for input?
Please help, thank you.

Code:

---

#include <stdio.h>

main()
{
        float cel, fah;
        char r;
        do
        {
                printf("Input the Fahrenheit temperature you want to convert: ");
                scanf("%f", &fah);
                cel = (5.0/9.0)*(fah - 32);
                printf("%5.1f degree F is %5.1f degree C. ", fah, cel);
                printf("Do you want to continue? ");
        }
        while ((r = getchar()) == 'y');
}

---

Because scanf() left the '\n' (newline) char still in the keyboard buffer (aka input stream in this case).

So when getchar() looks at the buffer, it see's the newline, pulls it off the buffer, and says "Tally Ho!". "I'm all set!".

put a getchar() after the scanf(), to pull the newline off the keyboard buffer, and all will be well.

Quote:

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Originally Posted by Adak

put a getchar() after the scanf(), to pull the newline off the keyboard buffer, and all will be well.

---

Unless of course the user inputted something like "51.6X" in which case the getchar() will retrieve the "X" and the newline still sits in the buffer, creating the same problem all over again.

There's no substitute for input validation.

Quote:

---

There's no substitute for input validation.

---

What is "input validation"?

And here is another exercise that I don't know how to solve:
Write a program to prove that "something == EOF" has only 2 values.

Here is my answer, it didn't work out, though.

Code:

---

#include <stdio.h>

main()
{
        char r;
        int i;

        do
        {
                i = (getchar() == EOF);
                printf("%d\n", i);
                getchar();
        }
        while ((r = getchar()) == 'y');
}

---

Please help me out.
Thank you.

that would never work.

simply do:

Code:

---

#include <stdio.h>

int main()
{
    printf("%d\n", EOF);
    getchar(); //pause program
    return 0;
}

---

That only prints out EOF.
And could you please tell me why that will not work?

Quote:

---

Originally Posted by kindlychung

What is "input validation"?

And here is another exercise that I don't know how to solve:
Write a program to prove that "something == EOF" has only 2 values.

Here is my answer, it didn't work out, though.

Code:

---

#include <stdio.h>

main()
{
        char r;
        int i;

        do
        {
                i = (getchar() == EOF);
                printf("%d\n", i);
                getchar();
        }
        while ((r = getchar()) == 'y');
}

---

Please help me out.
Thank you.

---

It looks about right - what are you seeing, and how do you "feed" your program?

--
Mats

https://apps.sourceforge.net/mediawi...n_input_buffer

Look at the example.

Quote:

---

Originally Posted by Elysia

https://apps.sourceforge.net/mediawi...n_input_buffer

Look at the example.

---

Authorization Required

This server could not verify that you are authorized to access the document requested. Either you supplied the wrong credentials (e.g., bad password), or your browser doesn't understand how to supply the credentials required.

Quote:

---

It looks about right - what are you seeing, and how do you "feed" your program?

---

It will give the value once and then terminate, in spite of the do...while... loop.

Quote:

---

Originally Posted by kindlychung

It will give the value once and then terminate, in spite of the do...while... loop.

---

Ah, yes, of course. If you make it terminate when you hit 'n' (or when you get back EOF?) rather than terminate on anything but 'y', it may work better.

--
Mats

Code:

---

                i = (getchar() == EOF);
                printf("%d\n", i);
               
        }
        while ((r = getchar()) != 'n');
}

---

mats,

I just changed it a little bit, and it seems to work. But I guess "!='n'" and "=='y'" are essentially the same thing, aka the condition of the loop, why are the effects so different?

Quote:

---

Originally Posted by kindlychung

Code:

---

                i = (getchar() == EOF);
                printf("%d\n", i);
               
        }
        while ((r = getchar()) != 'n');
}

---

mats,

I just changed it a little bit, and it seems to work. But I guess "!='n'" and "=='y'" are essentially the same thing, aka the condition of the loop, why are the effects so different?

---

because when you do '==y' you are only accepting the character 'y', and since when the user hits 'enter', it inputs a '\n' to the stream, your program will exit right after the first character you entered. what you really should be doing for your program to work is something like "=='y' || == '\n'"

because when you type 'y' in the console and hit enter, your input is actually: "y\n"

Quote:

---

Originally Posted by kindlychung

Code:

---

                i = (getchar() == EOF);
                printf("%d\n", i);
               
        }
        while ((r = getchar()) != 'n');
}

---

mats,

I just changed it a little bit, and it seems to work. But I guess "!='n'" and "=='y'" are essentially the same thing, aka the condition of the loop, why are the effects so different?

---

abraham sort of answers this, but I would put it this way: Your initial example exits whenever the reply is NOT 'y' - so anything OTHER than 'y' at that point will exit, the new code exits ONLY when the reply is 'n' - all other replies (including 'y', newline, etc) continues.

--
Mats

Quote:

---

Originally Posted by kindlychung

Authorization Required

This server could not verify that you are authorized to access the document requested. Either you supplied the wrong credentials (e.g., bad password), or your browser doesn't understand how to supply the credentials required.

---

Dang thing.
Strip the "s" in https:
http://apps.sourceforge.net/mediawik...n_input_buffer

I swear it. This new server is an annoying one.

Quote:

---

Originally Posted by kindlychung

But I guess "!='n'" and "=='y'" are essentially the same thing, aka the condition of the loop, why are the effects so different?

---

To you they are the same, because you're a good user and if presented with a y/n prompt, you enter either y or n. However there are tons of stupid users out there and they will enter "no" or "yes" or "blah".
What you want to make sure is that a user actually enters 'y' and that anything else is invalid. so you'd want to check that "== 'y'" (and if you allow a default yes check for '\n' as well).
As a rule of thumb, assume that your user doesn't know how your software works and assume they will do the dumbest things possible. Once you've anticipated the majority of these issues, your program will be pretty safe.

QuantumPete